Question

11．不等式組$\left\{\begin{array}{l}{|x|=x}\\{\frac{3-x}{3+x}＞|\frac{2-x}{2+x}|}\end{array}\right.$的解集是[0，$\sqrt{6}$]．

Answer 1

分析 把所給的絕對值不等式分類討論，等價轉化為$\left\{\begin{array}{l}{0≤x＜2}\\{\frac{3-x}{x+3}＞\frac{2-x}{x+2}}\end{array}\right.$ ①，或 $\left\{\begin{array}{l}{x≥2}\\{\frac{3-x}{x+3}＞\frac{x-2}{x+2}}\end{array}\right.$ ②．在分別求得①和②的解集，再取并集，即得所求．

解答 解：不等式組$\left\{\begin{array}{l}{|x|=x}\\{\frac{3-x}{3+x}＞|\frac{2-x}{2+x}|}\end{array}\right.$，即$\left\{\begin{array}{l}{x≥0}\\{\frac{3-x}{x+3}＞\frac{|x-2|}{x+2}}\end{array}\right.$，即 $\left\{\begin{array}{l}{0≤x＜2}\\{\frac{3-x}{x+3}＞\frac{2-x}{x+2}}\end{array}\right.$ ①，或 $\left\{\begin{array}{l}{x≥2}\\{\frac{3-x}{x+3}＞\frac{x-2}{x+2}}\end{array}\right.$ ②．
解①可得0≤x＜2，解②可得 2≤x＜$\sqrt{6}$，
綜上可得，不等式的解集為[0，$\sqrt{6}$]，
故答案為：[0，$\sqrt{6}$]．

點評 本題主要考查絕對值不等式的解法，體現(xiàn)了等價轉化和分類討論的數(shù)學思想，屬于中檔題．